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Chemical Reactions and Stoichiometry

Chemical Reactions and Stoichiometry. Chemical Stoichiometry. Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions. Atomic Masses. Elements occur in nature as mixtures of isotopes Carbon = 98.89% 12 C 1.11% 13 C <0.01% 14 C

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Chemical Reactions and Stoichiometry

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  1. Chemical Reactions and Stoichiometry

  2. Chemical Stoichiometry • Stoichiometry- The study of quantities of materials consumed and produced in chemical reactions.

  3. Atomic Masses • Elements occur in nature as mixtures of isotopes • Carbon = 98.89% 12C • 1.11% 13C • <0.01% 14C • Carbon atomic mass = 12.01 amu

  4. Example of Atomic Weight • 63 Cu 69.09% 65 Cu 30.91% • 63 Cu=62.93 x .6909 = 43.48 amu • 65Cu=64.93 x .3091 = 20.07 amu • Average Cu = 63.55 amu

  5. The Mole • The number equal to the number of carbon atoms in exactly 12 grams of pure 12C. • 1 moleof anything = 6.022  1023units of that thing

  6. Avogadro’s number equals6.022  1023 units

  7. Molar Mass • A substance’s molar mass(molecular weight) is the mass in grams of one mole of the compound. • CO2 = 44.01 grams per mole

  8. Calculating Molar Masses

  9. Calculations with Moles • Moles x MM = grams of substance • 1mole= 6.02x1023 particles= MM in grams • 23.5 g NaCl = ? Mol • 23.5 g / 58.5 MM = 0.402 mol • 2.4 mol CaCO3 = ? Grams • 2.4 mol x 100.0 MM = 240 grams • Use dimensional analysis bridge

  10. Mole Calculations

  11. Homework!! • p.117ff 6, 8, 13, 17c, 22a, 23, 26, 29, 32

  12. Percent Composition • Mass percent of an element: • For iron in iron (III) oxide, (Fe2O3)

  13. Sample % Composition • C10H14O --carvone-from carraway seeds • C 10 x 12.0 = 120g --> 120/150= 80% • H 14 x 1.0 = 14g --> 14/150= 9.3% • O 1 x 16.0 = 16g --> 16/150=10.7% • Total 150 g

  14. Another % Composition • C14H20N2SO4 --pennicillin F • C 14 x 12.0= 168 H 20 x 1.0 = 20.0 • N 2 x 14.0= 28.0 S 1 x 32.1 = 32.1 • O 4 x 16.0 = 64.0 Total= 312.1 • C=168/312.1=53.8% H=20/312.1=6.4% • N=28.0/312.1=9.0% S=32.1/312.1=10.3% • O=64.0/312.1=20.5%

  15. Formulas • molecular formula = (empirical formula)n [n = integer] • molecular formula = C6H6 = (CH)6 • empirical formula = CH

  16. Empirical Formula Determination From Percents • 1. Base calculation on 100 grams of compound. • 2. Determine moles of each element in 100 grams of compound. • 3. Divide each value of moles by the smallest of the values. • 4. Multiply each number by an integer to obtain all whole numbers.

  17. Sample Empirical Formula • 43.64% P 56.36% O • 43.64g | 1mol = 1.41 56.36g | 1mol =3.52 • | 31.0g | 16.0g • 1.41/1.41=1.0 3.52/1.41=2.50 • Multiply by 2 to fix the Oxygen • P=2 O=5 therefore P2O5 is the formula

  18. From Grams Found in Experiment • Cl 71.65g/ 35.5 =2.02 H 4.07g / 1.0 = 4.07 • C 24.27g / 12.0 = 2.02 • C= 2.02/2.02 =1 Cl= 2.02/2.02 =1 • H= 4.07/2.02=2 • Therefore H2CCl is the formula

  19. From Empirical Data • 0.1156 g of compound burned makes 0.1676 g H2O and 0.1638g CO2 • C= 0.1638 CO2 | 12 g C = .045g C • | 44g CO2 • H = 0.1676 H2O | 2g H = .019g H • | 18g H2O • O = .1156 - (.045+.019) = .052g O

  20. Continued • C= .045/12.0 = 0.00375 H=.019/1.0= 0.019 • O= .052 / 16.0=0.00325 • C= .00375/.00325 = 1.15 • H = .019 / .00325 = 5.84 • O= .00325/.00325=1.0 • Multiply by 7 (.15 is about one-seventh) • C8H41O7

  21. Molecular Formula • If molar mass is given, empirical can be adjusted to make molecular formula • From a previous slide, Cl=71.65% C=24.27% • H=4.07% Empirical formula was H2CCl • If MM given as 98.96, divide by Empirical mass (EM) to get the Multiplier • EM = 2.0+12.0+35.5 = 49.5 • MM/EM= 98.96/49.5 = 2 • Therefore formula is 2 x H2CCl = H4C2Cl2

  22. Homework!! • p. 119ff 35, 40, 43, 48, 54, • 57, 58

  23. Chemical Equations Chemical change involves a reorganization of the atoms in one or more substances. Algebraic combination of substances Reactants on left, Products on right.

  24. Chemical Equation • A representation of a chemical reaction: • C2H5OH + 3O22CO2 + 3H2O • reactants products

  25. Chemical Equation • C2H5OH +3O22CO2+3H2O • The equation is balanced, to obey Law of Conservation of Mass • 1 mole of ethanol reacts with 3 moles of oxygento produce • 2 moles of carbon dioxide and 3 moles of water

  26. Balancing Reactions • C6H14 + O2 CO2 + H2O • Make sure substances are written correctly • NEVER change formulas, only coefficients • C6H14 + O2 6 CO2 + 7 H2O • C6H14 + 19/2O2 6 CO2 + 7 H2O • This balances, but is not proper form, so multiply the whole by 2 • 2 C6H14 + 19 O2 12 CO2 + 14 H2O

  27. Balancing Reactions • (NH4)2Cr2O7 Cr2O3 + N2 + H2O • Start with the most complex substance • (NH4)2Cr2O7 Cr2O3 + N2 + 4 H2O • NH3 + O2 NO + H2O • 2 NH3 + 5/2O22 NO + 3 H2O • 4 NH3 + 5 O24 NO + 6 H2O

  28. Homework!! • p.121 61, 62, 63, 64, 65, 66

  29. Calculating Masses of Reactants and Products • 1. Balance the equation. • 2. Set up the mole bridge. • 3. Calculate moles of desired substance. • 4. Convert moles to grams, if required from the problem.

  30. Stoichiometric Problem Both baking soda, NaHCO3 and magnesium hydroxide, Mg(OH)2 are used as antacids. Which is more effective per gram in neutralizing stomach acid, HCl? NaHCO3 (s)+ HCl(aq) --> H2O(l) + CO2 (g)+ NaCl(aq) Mg(OH)2 (s)+ 2 HCl(aq) --> 2 H2O(l) + MgCl2 (aq) 1st equation 1 g NaHCO3 | 1 mol NaHCO3 | 1 mol HCl = 0.0119 mol | 84.0 g NaHCO3| 1 mol NaHCO3 2nd equation 1 g Mg(OH)2 | 1 mol Mg(OH)2 | 2 mol HCl = .0342 mol | 58.3 g Mg(OH)2 | 1 mol Mg(OH)2

  31. Another Normal Problem • Lithium hydroxide is used in spacecraft to eliminate CO2. How much CO2 is absorbed by 1.00 kg of LiOH? • LiOH(s) + CO2 (g) --> Li2CO3 (s) + H2O (l) • 2 LiOH(s) + CO2 (g) --> Li2CO3 (s) + H2O (l) • 1000 g LiOH | 1 mol LiOH | 1mol CO2 | 44 g CO2 • | 23.9 g LiOH | 2 mol LiOH | 1 mol CO2 • = 920. g CO2

  32. Limiting Reactant The limiting reactantis the reactant that is consumed first,limiting the amounts of products formed.

  33. Limiting Reactant Problem • The Haber process is used to make ammonia from hydrogen and nitrogen. If 25.0 kg of nitrogen and 5.00 kg hydrogen are mixed, how much ammonia is formed? • N2 (g) + 3 H2 (g) --> 2 NH3 (g) • Figure the limiting first: • 25000 g N2 | 1mol N2 | = 893 • | 28.0 g N2 | 1 mol N2 • 5000 g H2 | 1mol H2 | = 833.3 • | 2.0 g H2 | 3 mol H2 H2 is limiting

  34. Continue the problem with the limiting reactant: 5000 g H2 | 1mol H2 | 2 mol NH3| 17.0 g NH3 | 2.0 g H2 | 3 mol H2 | 1mol NH3 =28000 g NH3

  35. Nitrogen gas can be made by passing ammonia over copper(II) oxide at high temperatures. How much nitrogen will be made when 18.1 g NH3 is passed over 90.4 g of CuO, if Cu and H2O are the other products? 2 NH3 (g) + 3 CuO (s) --> N2 (g) + 3 Cu (s) + 3 H2O (g) 18.1 g NH3 | 1mol NH3 | = 0.532 | 17.0 g NH3 | 2 mol NH3 90.4 g CuO | 1 mol CuO | | = 0.379 | 79.5 g CuO | 3 mol CuO CuO is limiting 90.4 g CuO | 1 mol CuO | 1 mol N2 | 28.0 g N2=10.6 g N2 | 79.5 g CuO | 3 mol CuO | 1 mol N2 Limiting Reactant 2

  36. Solving a Stoichiometry Problem • 1. Balance the equation. • 2. Convert masses to moles. • 3. Determine which reactant is limiting. • 4. Use moles of limiting reactant and mole bridge to find moles of desired product. • 5. Convert from moles to grams.

  37. Percent Yield • You never get the full or Theoretical amount from a chemical reaction, due to errors and other circumstances. • Percent yield is the percent of the theoretical you did receive • % yield = actual yield • theoretical yield

  38. Application of % yield • From a previous slide, if the actual yield of Nitrogen from ammonia was 8.75 g, what was the percent yield? • % = actual = 8.75 = 82.5 % • theoretical 10.6 • From another, if the percent yield of ammonia were 24.7 %, what was the actual yield? • Actual = theoretical x % • 28000 g x 0.247 = 6916 g --> 6900 g

  39. Homework!! • p. 121 ff 69, 70, 76, 81, 82, 84, 90, 95, 108 • SUPERXCR!!! 112

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